Exam 1 Solutions February 3, 2000 6

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Exam 1 Solutions
February 3, 2000
ECE 221: Electric Circuits
Dr. McNames
• Write your 6-digit identification number and student
identification numbers below.
• Do not begin the exam or look at the problems until instructed
to do so.
• You have 100 minutes to complete the exam.
• Once you begin, write your student ID at the top of each page.
• Do not use separate scratch paper. If you need more space, use
the backs of the exam pages.
• If you have extra time, double check your answers. If you run
out of time, write a note describing your strategy and equations
that can be used to help solve the problem.
Problem 1:______ /
Problem 2:______ /
Problem 3:______ /
Problem 4:______ /
17
20
13
10
Total:______ / 60
6-Digit Identification Number:_____________
Student Identification Number:_____________
Student ID:_________________
2 of 6
1. Mesh Current Method (17 points)
10 Ω
ia
2v 1
50 Ω
20 V
ic
60 Ω
ib
80 Ω
40 Ω
v1
a. (9 pts) For the circuit shown above, write the mesh-current equations around each
loop in terms of the currents ia , ib , and ic. You do not need to simplify your equations.
Loop ia : 10i a + 2 ⋅ 40i c + 50(i a − i c ) = 0
Loop ib : − 2 ⋅ 40i c + 60ib + 80(ib − ic ) = 0
Loop ic : 50(i c − ia ) + 80( ic − ib ) + 40ic = 20
b. (3 pts) Solve for the currents ia , ib , and ic.
ia = -96.6 mA
ib = 221 mA
ic = 193 mA
c. (2 pts) How much power is being delivered by the independent source?
P20V = 20 ic = 3.86 W
d. (2 pts) How much power is being delivered by the dependent source?
P2V1 = 2 ⋅ v1 (ib − ia ) = 2 ⋅ 40ic ( ib − i a ) = 4.90 W
e. (1 pt) How much power is being absorbed by the resistors?
PR = P20V + P2V1 = 8.76 W
Student ID:_________________
3 of 6
2. Node Voltage Method (20 pts)
30iα
i14
4Ω
1
50Ω
2
6Ω
2Ω
3
iα
v1
20 V
40 Ω
v3
20 Ω
4
v4
a. (6 pts) For the circuit shown above, write the node-voltage equations at each of the
specified nodes in terms of v 1 , v3 , v 4 , and i14 . You do not need to simplify your
equations.
Node 1:
v1 v1 − 20
+
+ i14 = 0
50
4
Node 3:
v3 − 20 v3 v3 − v4
+
+
=0
6
40
2
Node 4:
v4 − v3 v 4
+
− i14 = 0
2
20
b. (1 pt) Write an expression for iα in terms of v 3 and v 4 .
iα= v4 − v3
2
c. (2 pts) In part a. you should have written 3 independent equations in 4 unknowns.
Write a fourth independent equation that relates the voltages between nodes 1 and 4
to the voltages v 3 and v 4 . Hint: use your answer from part b.
Eq:
30( v4 − v3 )
= v4 − v1
2
Student ID:_________________
4 of 6
2. Node Voltage Method – Continued
The same circuit from the previous page is shown below for convenience.
30iα
i14
4Ω
1
50Ω
2
6Ω
2Ω
3
iα
v1
20 V
40 Ω
v3
20 Ω
4
v4
d. (6 pts) Use a supernode to write three independent equations in terms of v 1 , v3 , and v 4 .
You do not need to simplify your equations.
Eq. 1:
v1 v1 − 20 v4 − v3 v 4
+
+
+
=0
50
4
2
20
Eq. 2:
v3 − 20 v3 v3 − v4
+
+
=0
6
40
2
Eq. 3:
30( v4 − v3 )
= v4 − v1
2
e. (3 pts) Solve for the voltages v 1 , v3 , and v 4 .
v1 = 14.7 V
v3 = 18.0 V
v4 = 18.2 V
f. (2 pts) How much power is being delivered by the dependent source?
v4 − v3
=115 mA
2
v
P30 = 30iα ⋅ i14 = 30iα  iα + 4  =3.54 W
20 

iα =
Student ID:_________________
5 of 6
3. Thevenin/Norton Equivalents & Maximum Power Transfer (13 pts)
Hint: think carefully about how to approach this problem. Some approaches are easier
than others.
a
b
8 kΩ
20 kΩ
40V
30 kΩ
2 mA
5 kΩ
a. (4 pts) Find the current flowing from the node labeled a to the node labeled b if the
nodes are connected together (short circuit current).
Solve part b. first, then calculate Isc by the relationship of Req to Voc.
Req = (20 k || 30k ) + 8k = 20 kΩ
ISC =
VOC
= -1.2 mA
Req
b. (4 pts) Find the voltage between the terminals a and b if the nodes are left
unconnected (open circuit voltage), as shown.
Label the voltage across the 30k Ohm resistor as V and apply the node-voltage
method.
V − 40 V
+
− 2m = 0 ; V = 48 V
20 k
30k
VOC= 40 − V − 2m ⋅ 8k = -24 V
c. (4 pts) Draw the Thevenin and Norton equivalents of the circuit as seen from the
nodes a and b. Clearly label these nodes.
20 kΩ
a
24 V
a
1.2 mA
b
20 kΩ
b
d. (1 pt) Suppose a resistor RL is connected to the nodes a and b. What value of RL will
maximize the power delivered to RL ?
RL = 20 kΩ
Student ID:_________________
6 of 6
4. Operational Amplifiers (10 pts)
For both circuits below, assume that the operational amplifier is ideal and operating in the
linear region.
a. (4 pts) Find the values of Rf and Rb so that v o = 20ib – 10ia .
Rf
Ra
ia
vo
RL
Rb
ib
v p = ib Rb
vo − v p
Rf
= −i a
vo = Rbi b − R f ia
Rb = 20 Ω
Rf = 10 Ω
b. (6 pts) Find an expression for v o as a function of v 1 and i2 .
R3
R1
R2
v1
R4
vf
i2
v f = − R4 i2
vf
v −v
− R4 i2
= o 1 =
R3 R3 + R1
R3
 R + R1 
 R4i 2
vo = v1 −  3
 R3 
RL
vo